Random variables

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Show that The standard deviation is zero if and only if X is a constant function,that is ,X(s) = k for every s belonging to S,or ,simply X=k.

When they say constant function it means every element in S is been mapped to single element in the range.That is the single element is k.
Which means all random variables are taking a single value which is K.
I don't if I should assume that it is an equiprobable space or no.
Due to that I'm getting confused in calculating my mean n thus heading towards showing std dev 0
I don't know if I'm thinking in the right direction
 

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  • #2
andrewkirk
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A bit more information is needed about the problem, in particular whether the sample space S contains only finitely many elements, in which case we say that X is a discrete random variable. The statement you are asked to prove is not true for a non-discrete (aka 'continuous') random variable, but there is a similar statement which is true, which is that the standard deviation is zero if and only if the random variable is 'almost surely' constant. That means that there exists some value k such that Prob(X(s)=k)=1.

So let's assume S has only finitely many elements. You do not need to assume anything about equiprobability in order to prove the statement. All you need to assume is that no element s of the sample space has probability zero, which is an assumption that is usually implicitly made for any discrete random variable.

I suggest you start by assuming S has n elements ##s_1,...,s_n##, with values ##X_1,...,X_n## and probabilities ##p_1,...,p_n## (which must add to 1, and none of which are zero) and use that to write out an expression for the standard deviation. Then assume there is at least one element, say ##X_k## that is not equal to the mean ##\bar X## and see if you can prove that the standard deviation must be nonzero.
 
  • #3
FactChecker
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Show that The standard deviation is zero if and only if X is a constant function,that is ,X(s) = k for every s belonging to S,or ,simply X=k.

When they say constant function it means every element in S is been mapped to single element in the range.That is the single element is k.
Which means all random variables are taking a single value which is K.
I don't if I should assume that it is an equiprobable space or no.
No. You should not assume that.
Due to that I'm getting confused in calculating my mean n thus heading towards showing std dev 0
If the variable is a constant, k, what can you say about k and the mean, ##\mu##?
I don't know if I'm thinking in the right direction
For a homework problem, you need to show work and use the homework template. Then people can tell you if you are going in the right direction. Your profile says that you have a degree in statistics, so you should be able to show a corresponding level of work.
 
  • #4
StoneTemplePython
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my approach would be to work directly with the CDF of ##X## and the definition of strict convexity.

From there work backwards to why Jensen's Inequality gives you the insight you want regarding variance (and std deviation is zero iff variance is zero).

If needed, you can then also work backward to the sample space implications, supposing that you understand the links between CDFs and sample spaces. In the case of an uncountable sample space, the term "almost" will show up...

- - - -
I suppose my approach de-emphasizes the sample space, but in general convexity is a big deal, and CDFs are extremely useful.
 

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